Review question

# Can we find the area between $y=e^{-x}$ and $y=e^{-x}\sin x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9312

## Solution

The curves $C_1$ and $C_2$ are defined by $y=e^{-x}\qquad (x>0)\qquad \text{and}\qquad y=e^{-x}\sin x \qquad (x>0),$ respectively. Sketch roughly $C_1$ and $C_2$ on the same diagram.

For $C_1$ and $C_2$, we need the parts of these curves that satisfy $x > 0$.

The two curves $\,y=e^{-x}$ and $\,y=-e^{-x}$ form what is known as an `envelope’ around $\,y=e^{-x}\sin x$.

Let $x_n$ denote the $x$-coordinate of the $n$th point of contact between the two curves, where $0<x_1< x_2<\dotsb$, and let $A_n$ denote the area of the region enclosed by the two curves between $x_n$ and $x_{n+1}$. Show that $A_n=\frac{1}{2}\left(e^{2\pi}-1\right)e^{-(4n+1)\pi/2}$

First we need to know what the points $x_n$ are. Solving $e^{-x}=e^{-x}\sin x,\qquad \left(e^{-x}\neq 0 \right)$ $\Longrightarrow \sin x = 1 \qquad \Longrightarrow \qquad x_1=\frac{\pi}{2},\, x_2=2\pi+\frac{\pi}{2} \dotsb$ Thus $x_n=2\pi(n-1)+\frac{\pi}{2}=\frac{1}{2}(4n-3)\pi$ and $A_n=\int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi}\,e^{-x}\left(1-\sin x\right)\,dx,$ the difference between the two curves $C_1$ and $C_2$. This can be seen below.

Consider $I=\int e^{-x}\sin x\,dx$. We can solve this via integration by parts, with $u=\sin x$ and $v'=e^{-x}$: \begin{align*} I &= \int e^{-x}\sin x \,dx \\ &= \left[-e^{-x}\sin x \right] + \int e^{-x}\cos x \,dx \\ &= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - \int e^{-x}\sin x \,dx \, \, \text{ (by integrating the integral above by parts again)}\\ &= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - I\\ \Longrightarrow \quad 2I &= -e^{-x}\left(\sin x+ \cos x\right)\\ \Longrightarrow \quad I &= -\frac{1}{2}e^{-x}\left(\sin x+ \cos x\right) \end{align*} Thus \begin{align*} A_n &=\int_{x_n}^{x_{n+1}}\,e^{-x}\left(1-\sin x\right)\,dx \\ &= \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\,dx - \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\sin x\,dx \\ &= \left[-e^{-x} + \frac{1}{2}e^{-x}\left(\sin x+ \cos x\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \, \, \text{(recognising the second integral above as I)}\\ &= \left[\frac{1}{2}e^{-x}\left(\sin x+ \cos x-2\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \\ &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left[\sin \left(2n\pi+\frac{\pi}{2}\right) + \cos \left(2n\pi+\frac{\pi}{2}\right) -2\right] \\ &\quad - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left[\sin \left(2(n-1)\pi+\frac{\pi}{2}\right) + \cos \left(2(n-1)\pi+\frac{\pi}{2}\right) -2\right]. \end{align*} With $\,\sin\left(2n\pi + \frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1$ and $\,\cos\left(2n\pi + \frac{\pi}{2}\right)=\cos\frac{\pi}{2}=0$ for all integers $n$, we find \begin{align*} A_n &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(1 + 0 -2 \right) - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left(1 + 0 -2\right) \\ &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(-1 +e^{2\pi} \right) \\ &= \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi}, \end{align*}

as required.

[…] and hence find $\displaystyle\sum\limits_{n=1}^{\infty} A_n$.

Let $\,A=\displaystyle\sum\limits_{n=1}^{\infty} A_n$. Then \begin{align*} A &= \sum\limits_{n=1}^{\infty} \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi} \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) \sum\limits_{n=1}^{\infty} e^{-2n\pi}e^{-\frac{\pi}{2}} \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \sum\limits_{n=1}^{\infty} (e^{-2\pi})^n \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \left[e^{-2\pi} + (e^{-2\pi})^2 + (e^{-2\pi})^3 + \dotsb\right]\\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} e^{-2\pi} \left[1+ e^{-2\pi} + (e^{-2\pi})^2 + \dotsb\right] \end{align*} We therefore have the infinite sum of a geometric series, with first term $\,a=\frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}$ and ratio $\,r=e^{-2\pi}$. Since $\,2\pi > 0$, $e^{-2\pi}<1$ and the series converges, with infinite sum $\,S_{\infty}=\frac{a}{1-r}$. Therefore \begin{align*} A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}\times \frac{1}{1-e^{-2\pi}} \\ &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}} \times \frac{e^{2\pi}}{e^{2\pi}-1} \\ &= \frac{1}{2}e^{-\frac{5\pi}{2}} \times e^{2\pi} = \frac{1}{2}e^{-\frac{\pi}{2}}. \end{align*}

So $\displaystyle\sum\limits_{n=1}^{\infty} A_n = \frac{1}{2}e^{-\frac{\pi}{2}}$.